JEE CHEMISTRY MCQ: CO-ORDINATION COMPOUNDS

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Analyzing +4 | -1

A s- block element (M) reacts with oxygen to form an oxide of the formula

M{O_2}.

The oxide is pale yellow in colour and paramagnetic. The element(M) is

View answer and explanation

Correct option: D

Option D (K) is correct because potassium reacts with oxygen to form potassium superoxide (

KO_2

), which is paramagnetic due to the presence of an unpaired electron.

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\mathrm{K}\left[ {\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}} \right]

\left[ {\mathrm{PtC}{\mathrm{l}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\mathrm{B}{\mathrm{r}_2}\mathrm{\;and\;}\left[ {\mathrm{PtB}{\mathrm{r}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\mathrm{C}{\mathrm{l}_2}\mathrm{\;}

\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]\left[\mathrm{PtCl}_4\right]

\left[{Co}\left({NH}_3\right)_5\left({NO}_2\right)\right] {Cl}_2

\left[{Co}\left({NH}_3\right)_5({ONO})\right] {Cl}_2 \ldots \ldots \ldots

\left[{Cu}\left({NH}_3\right)_4\right]\left[{PtCl}_4\right] \,\, \& \,\, \left[{Pt}\left({NH}_3\right)_4\right]\left[{CuCl}_4\right] \ldots \ldots \ldots

\left[{Pt}\left({NH}_3\right)_4 {Cl}_2\right] {Br}_2 \, \, \& \, \, \left[{Pt}\left({NH}_3\right)_4 {Br}_2\right]{Cl}_2 \ldots \ldots \ldots